Thickness of the subgroup intersection graph of a finite group

Huadong Su; Ling Zhu; Huadong Su; Ling Zhu

doi:10.3934/math.2021157

AIMS Mathematics

2021, Volume 6, Issue 3: 2590-2606. doi: 10.3934/math.2021157

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Research article

Thickness of the subgroup intersection graph of a finite group

Huadong Su ^{1
,
,},
Ling Zhu ²

1.
School of Science, Beibu Gulf University, Qinzhou, Guangxi, 535011, P. R. China
2.
School of Artificial Intelligence, Jiangxi University of Applied Science, Nanchang, Jiangxi, 330100, P.R, China

Received: 11 September 2020 Accepted: 21 December 2020 Published: 25 December 2020
MSC : 05C25, 05C10, 20D60

Let $G$ be a finite group. The intersection graph of subgroups of $G$ is a graph whose vertices are all non-trivial subgroups of $G$ and in which two distinct vertices $H$ and $K$ are adjacent if and only if $H\cap K\neq 1$ . In this paper, we classify all finite abelian groups whose thickness and outerthickness of subgroup intersection graphs are 1 and 2, respectively. We also investigate the thickness and outerthickness of subgroup intersection graphs for some finite non-abelian groups.

Keywords:

Citation: Huadong Su, Ling Zhu. Thickness of the subgroup intersection graph of a finite group[J]. AIMS Mathematics, 2021, 6(3): 2590-2606. doi: 10.3934/math.2021157

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Abstract

1. Introduction and preliminaries

Let $W$ be a set and $H:W\longrightarrow W$ be a mapping. A point $w\in W$ is called a fixed point of $H\$ if $w = Hw.$ Fixed point theory plays a fundamental role in functional analysis (see ^[15]). Shoaib ^[17] introduced the concept of $\alpha$ -dominated mapping and obtained some fixed point results (see also ^[1,2]). George et al. ^[11] introduced a new space and called it rectangular $b$ -metric space ( $r.b.m.$ space). The triangle inequality in the $b$ -metric space was replaced by rectangle inequality. Useful results on $r.b.m.$ spaces can be seen in (^[5,6,8,9,10]). Ćirić introduced new types of contraction and proved some metrical fixed point results (see ^[4]). In this article, we introduce Ćirić type rational contractions for $\alpha$ -dominated mappings in $r.b.m.$ spaces and proved some metrical fixed point results. New interesting results in metric spaces, rectangular metric spaces and $b$ -metric spaces can be obtained as applications of our results.

Definition 1.1. ^[11] Let $U$ be a nonempty set. A function $d_{lb}:U\times U\rightarrow \lbrack 0, \infty)$ is said to be a rectangular $b$ -metric if there exists $b\geq 1$ such that

(ⅰ) $d_{lb}(\theta, \nu) = d_{lb}(\nu, \theta$ );

(ⅱ) $d_{lb}(\theta, \nu) = 0$ if and only if $\theta = \nu;$

(ⅲ) $d_{lb}(\theta, \nu)\leq b[d_{lb}(\theta, q)+d_{lb}(q, l)+d_{lb}(l, \nu)]$ for all $\theta, \nu \in U$ and all distinct points $q, l\in U \backslash \{ \theta, \nu \}.$

The pair $(U, d_{lb})$ is said a rectangular $b$ -metric space (in short, $r.b.m.$ space) with coefficient $b$ .

Definition 1.2. ^[11] Let $(U, d_{lb})$ be an $r.b.m.$ space with coefficient $b$ .

(ⅰ) A sequence $\{ \theta_{n}\}$ in $(U, d_{lb})$ is said to be Cauchy sequence if for each $\varepsilon > 0$ , there corresponds $n_{0}\in \mathbb{N}$ such that for all $n, m\geq n_{0}$ we have $d_{lb}(\theta_{m}, \theta_{n}) < \varepsilon$ or $\lim_{n, m\rightarrow +\infty }d_{lb}(\theta_{n}, \theta_{m}) = 0.$

(ⅱ) A sequence $\{ \theta_{n}\}$ is rectangular $b$ -convergent (for short, ( $d_{lb}$ )-converges) to $\theta$ if $\lim_{n\rightarrow +\infty }d_{lb}(\theta _{n}, \theta) = 0.$ In this case $\theta$ is called a ( $d_{lb}$ )-limit of $\{ \theta _{n}\}.$

(ⅲ) $(U, d_{lb})$ is complete if every Cauchy sequence in $U \; d_{lb}$ -converges to a point $\theta \in U$ .

Let $\varpi _{b}$ , where $b\geq 1,$ denote the family of all nondecreasing functions $\delta _{b}:[0, +\infty)\rightarrow \lbrack 0, +\infty)$ such that $\sum_{k = 1}^{+\infty }b^{k}\delta _{b}^{k}(t) < +\infty$ and $b\delta _{b}(t) < t$ for all $t > 0,$ where $\delta _{b}^{k}$ is the $k^{th}$ iterate of $\delta _{b}.$ Also $b^{n+1}\delta _{b}^{n+1}(t) = b^{n}b\delta _{b}(\delta _{b}^{n}(t)) < b^{n}\delta _{b}^{n}(t).$

Example 1.3. ^[11] Let $U \; = \mathbb{N}.$ Define $d_{lb}:U\times U\rightarrow \mathbb{R} ^{+}\cup \{0\}$ such that $d_{lb}(u, v) = d_{lb}(v, u)$ for all $u, v\in U$ and $\alpha > 0$

$\begin{equation*} d_{lb}(u,v) = \left\{ \begin{array}{c} 0,\text{ if }u = v; \\ 10\alpha ,\text{ if }u = 1,\text{ }v = 2; \\ \alpha ,\text{ if }u\in \{1,2\}\text{ and }v\in \{3\}; \\ 2\alpha ,\text{ if }u\in \{1,2,3\}\text{ and }v\in \{4\}; \\ 3\alpha ,\text{ if }u\text{ or }v\notin \{1,2,3,4\}\text{ and }u\neq v. \end{array} \right. \end{equation*}$

Then $(U, d_{lb})$ is an $r.b.m.$ space with $b = 2 > 1.$ Note that

$\begin{equation*} d(1,4)+d(4,3)+d(3,2) = 5\alpha \lt 10\alpha = d(1,2). \end{equation*}$

Thus $d_{lb}$ is not a rectangular metric.

Definition 1.4. ^[17] Let $(U, d_{lb})$ be an $r.b.m.$ space with coefficient $b$ . Let $S:U\rightarrow U$ be a mapping and $\alpha :U\times U\rightarrow \lbrack 0, +\infty)$ . If $A\subseteq U,$ we say that the $S$ is $\alpha$ -dominated on $A,$ whenever $\alpha (i, Si)\geq 1$ for all $i\in A.$ If $A = U$ , we say that $S$ is $\alpha$ -dominated $.$

For $\theta, \nu \in U$ , $a > 0,$ we define $D_{lb}(\theta, \nu)$ as

$\begin{equation*} D_{lb}( \theta,\nu ) = \max \{d_{lb}( \theta,\nu ),\frac{d_{lb}\left( \theta,S \theta\right) .d_{lb}\left( \nu ,S\nu \right) }{ a+d_{lb}\left( \theta,\nu \right) },d_{lb}( \theta,S \theta ),d_{lb}(\nu ,S\nu )\}. \end{equation*}$

2. Main result

Now, we present our main result.

Theorem 2.1. Let $(U, d_{lb})$ be a complete $r.b.m.$ space with coefficient $b$ , $\alpha :U\times U\rightarrow \lbrack 0, \infty), \; S:U\rightarrow U$ , $\{ \theta_{n}\}$ be a Picard sequence and $S$ be a $\alpha$ -dominated mapping on $\{ \theta_{n}\}.$ Suppose that, for some $\delta _{b}\in \varpi _{b}$ , we have

$\begin{equation} d_{lb}(S \theta,S\nu )\leq \delta _{b}(D_{lb}( \theta,\nu )), \end{equation}$

(2.1)

for all $\theta, \nu \in \{ \theta_{n}\}$ with $\alpha (\theta, \nu)\geq 1$ . Then $\{ \theta_{n}\}$ converges to $\theta^{\ast }\in U.$ Also, if (2.1) holds for $\theta^{\ast }$ and $\alpha (\theta_{n}, \theta^{\ast })\geq 1$ for all $n\in \mathbb{N} \cup \{0\}$ , then $S$ has a fixed point $\theta^{\ast }$ in $U.$

Proof. Let $\theta_{0}\in U$ be arbitrary. Define the sequence $\{ \theta _{n}\}$ by $\theta_{n+1} = S \theta_{n}$ for all $n\in \mathbb{N} \cup \{0\}.$ We shall show that $\{ \theta_{n}\}$ is a Cauchy sequence. If $\theta _{n} = \theta_{n+1},$ for some $n\in \mathbb{N},$ then $\theta_{n}$ is a fixed point of $S$ . So, suppose that any two consecutive terms of the sequence are not equal. Since $S:U\rightarrow U$ be an $\alpha$ -dominated mapping on $\{ \theta_{n}\}$ , $\alpha (\theta _{n}, S \theta_{n})\geq 1$ for all $n\in \mathbb{N} \cup \{0\}$ and then $\alpha (\theta_{n}, \theta_{n+1})\geq 1$ for all $n\in \mathbb{N} \cup \{0\}.$ Now by using inequality (2.1), we obtain

$\begin{eqnarray*} d_{lb}( \theta_{n+1}, \theta_{n+2}) & = &d_{lb}(S \theta_{n},S \theta_{n+1})\leq \delta _{b}(D_{lb}( \theta_{n}, \theta_{n+1})) \\ &\leq &\delta _{b}(\max \{d_{lb}( \theta_{n}, \theta_{n+1}),\frac{ d_{lb}\left( \theta_{n}, \theta_{n+1}\right) .d_{lb}\left( \theta_{n+1}, \theta_{n+2}\right) }{a+d_{lb}\left( \theta_{n}, \theta_{n+1}\right) }, \\ &&d_{lb}( \theta_{n}, \theta_{n+1}),d_{lb}( \theta_{n+1}, \theta_{n+2})\}) \\ &\leq &\delta _{b}(\max \{d_{lb}( \theta_{n}, \theta_{n+1}),d_{lb}( \theta_{n+1}, \theta_{n+2})\}). \end{eqnarray*}$

If $\max \{d_{lb}(\theta_{n}, \theta_{n+1}), d_{lb}(\theta_{n+1}, \theta _{n+2})\} = d_{lb}(\theta_{n+1}, \theta_{n+2}),$ then

$\begin{eqnarray*} d_{lb}( \theta_{n+1}, \theta_{n+2}) &\leq &\delta _{b}(d_{lb}( \theta_{n+1}, \theta_{n+2})) \\ &\leq &b\delta _{b}(d_{lb}( \theta_{n+1}, \theta_{n+2})). \end{eqnarray*}$

This is the contradiction to the fact that $b\delta _{b}(t) < t$ for all $t > 0.$ So

$\begin{equation*} \max \{d_{lb}( \theta_{n}, \theta_{n+1}),d_{lb}( \theta_{n+1}, \theta_{n+2})\} = d_{lb}( \theta_{n}, \theta_{n+1}). \end{equation*}$

Hence, we obtain

$\begin{equation*} d_{lb}( \theta_{n+1}, \theta_{n+2})\leq \delta _{b}(d_{lb}( \theta_{n}, \theta_{n+1}))\leq \delta _{b}^{2}(d_{lb}( \theta_{n-1}, \theta_{n})) \end{equation*}$

Continuing in this way, we obtain

$\begin{equation} d_{lb}( \theta_{n+1}, \theta_{n+2})\leq \delta _{b}^{n+1}(d_{lb}( \theta_{0}, \theta_{1})). \end{equation}$

(2.2)

Suppose for some $n, m\in \mathbb{N}$ with $m > n,$ we have $\theta_{n} = \theta_{m}$ . Then by (2.2)

$\begin{eqnarray*} d_{lb}( \theta_{n}, \theta_{n+1}) & = &d_{lb}( \theta_{n},S \theta_{n}) = d_{lb}( \theta_{m},S \theta_{m}) = d_{lb}( \theta_{m}, \theta_{m+1}) \\ &\leq &\delta _{b}^{m-n}\left( d_{lb}( \theta_{n}, \theta _{n+1})\right) \lt b\delta _{b}\left( d_{lb}( \theta_{n}, \theta_{n+1})\right) \end{eqnarray*}$

As $d_{lb}(\theta_{n}, \theta_{n+1}) > 0,$ so this is not true, because $b\delta _{b}(t) < t$ for all $t > 0.$ Therefore, $\theta_{n}\neq \theta_{m}$ for any $n, m\in \mathbb{N}$ . Since $\sum_{k = 1}^{+\infty }b^{k}\delta _{b}^{k}(t) < +\infty,$ for some $\nu \in \mathbb{N},$ the series $\sum_{k = 1}^{+\infty }b^{k}\delta _{b}^{k}(\delta _{b}^{\nu -1}(d_{lb}(\theta_{0}, \theta_{1})))$ converges. As $b\delta _{b}(t) < t,$ so

$\begin{equation*} b^{n+1}\delta _{b}^{n+1}(\delta _{b}^{\nu -1}(d_{lb}( \theta_{0}, \theta_{1}))) \lt b^{n}\delta _{b}^{n}(\delta _{b}^{\nu -1}(d_{lb}( \theta _{0}, \theta_{1}))), \text{ for all }n\in \mathbb{N}. \end{equation*}$

Fix $\varepsilon > 0.$ Then $\frac{\varepsilon }{2} = \varepsilon ^{\prime } > 0.$ For $\varepsilon ^{\prime },$ there exists $\nu (\varepsilon ^{\prime })\in \mathbb{N}$ such that

$\begin{equation} b\delta _{b}(\delta _{b}^{\nu (\varepsilon ^{\prime })-1}(d_{lb}( \theta _{0}, \theta_{1})))+b^{2}\delta _{b}^{2}(\delta _{b}^{\nu (\varepsilon ^{\prime })-1}(d_{lb}( \theta_{0}, \theta_{1})))+\cdots \lt \varepsilon ^{\prime } \end{equation}$

(2.3)

Now, we suppose that any two terms of the sequence $\{ \theta_{n}\}$ are not equal. Let $n, m\in \mathbb{N}$ with $m > n > \nu (\varepsilon ^{\prime }).$ Now, if $m > n+2$ ,

$\begin{eqnarray*} d_{lb}( \theta_{n}, \theta_{m}) &\leq &b[d_{lb}( \theta_{n}, \theta_{n+1})+d_{lb}( \theta_{n+1}, \theta_{n+2})+d_{lb}( \theta_{n+2}, \theta_{m})] \\ &\leq &b[d_{lb}( \theta_{n}, \theta_{n+1})+d_{lb}( \theta_{n+1}, \theta_{n+2})]+b^{2}[d_{lb}( \theta_{n+2}, \theta_{n+3}) \\ &&+d_{lb}( \theta_{n+3}, \theta_{n+4})+d_{lb}( \theta_{n+4}, \theta_{m})] \\ &\leq &b[\delta _{b}^{n}\left( d_{lb}( \theta_{0}, \theta _{1})\right) +\delta _{b}^{n+1}\left( d_{lb}( \theta_{0}, \theta_{1})\right) ]+b^{2}[\delta _{b}^{n+2}\left( d_{lb}( \theta_{0}, \theta_{1})\right) \\ &&+\delta _{b}^{n+3}\left( d_{lb}( \theta_{0}, \theta_{1})\right) ]+b^{3}[\delta _{b}^{n+4}\left( d_{lb}( \theta_{0}, \theta _{1})\right) +\delta _{b}^{n+5}\left( d_{lb}( \theta_{0}, \theta_{1})\right) ]+\cdots \\ &\leq &b\delta _{b}^{n}(d_{lb}( \theta_{0}, \theta_{1}))+b^{2}\delta _{b}^{n+1}(d_{lb}( \theta_{0}, \theta_{1}))+b^{3}\delta _{b}^{n+2}\left( d_{lb}( \theta_{0}, \theta_{1})\right) +\cdots \\ & = &b\delta _{b}(\delta _{b}^{n-1}(d_{lb}( \theta_{0}, \theta_{1})))+b^{2}\delta _{b}^{2}(\delta _{b}^{n-1}(d_{lb}( \theta_{0}, \theta_{1})))+\cdots . \end{eqnarray*}$

By using (2.3), we have

$\begin{eqnarray*} &&d_{lb}( \theta_{n}, \theta_{m}) \\ & \lt &b\delta _{b}(\delta _{b}^{\nu (\varepsilon ^{\prime })-1}(d_{lb}( \theta_{0}, \theta_{1})))+b^{2}\delta _{b}^{2}(\delta _{b}^{\nu (\varepsilon ^{\prime })-1}(d_{lb}( \theta_{0}, \theta_{1})))+\cdots \lt \varepsilon ^{\prime } \lt \varepsilon . \end{eqnarray*}$

Now, if $m = n+2$ , then we obtain

$\begin{eqnarray*} &&d_{lb}( \theta_{n}, \theta_{n+2}) \\ &\leq &b[d_{lb}( \theta_{n}, \theta_{n+1})+d_{lb}( \theta_{n+1}, \theta_{n+3})+d_{lb}( \theta_{n+3}, \theta_{n+2})] \\ &\leq &b[d_{lb}( \theta_{n}, \theta_{n+1})+b[d_{lb}( \theta_{n+1}, \theta_{n+2})+d_{lb}( \theta_{n+2}, \theta_{n+4})+d_{lb}( \theta_{n+4}, \theta_{n+3})] \\ &&+d_{lb}( \theta_{n+3}, \theta_{n+2})] \\ &\leq &bd_{lb}( \theta_{n}, \theta_{n+1})+b^{2}d_{lb}( \theta_{n+1}, \theta_{n+2})+bd_{lb}( \theta_{n+2}, \theta_{n+3})+b^{2}d_{lb}( \theta_{n+3}, \theta_{n+4}) \\ &&+b^{3}[d_{lb}( \theta_{n+2}, \theta_{n+3})+d_{lb}( \theta_{n+3}, \theta_{n+5})+d_{lb}( \theta_{n+5}, \theta_{n+4})] \\ &\leq &bd_{lb}( \theta_{n}, \theta_{n+1})+b^{2}d_{lb}( \theta_{n+1}, \theta_{n+2})+\left( b+b^{3}\right) d_{lb}( \theta_{n+2}, \theta_{n+3})+b^{2}d_{lb}( \theta_{n+3}, \theta_{n+4}) \\ &&+b^{3}d_{lb}( \theta_{n+5}, \theta_{n+4})+b^{4}[d_{lb}( \theta_{n+3}, \theta_{n+4})+d_{lb}( \theta_{n+4}, \theta_{n+6})+d_{lb}( \theta_{n+6}, \theta_{n+5})] \\ &\leq &bd_{lb}( \theta_{n}, \theta_{n+1})+b^{2}d_{lb}( \theta_{n+1}, \theta_{n+2})+\left( b+b^{3}\right) d_{lb}( \theta_{n+2}, \theta_{n+3}) \\ &&+\left( b^{2}+b^{4}\right) d_{lb}( \theta_{n+3}, \theta_{n+4})+b^{3}d_{lb}( \theta_{n+5}, \theta_{n+4})+b^{4}d_{lb}( \theta_{n+6}, \theta_{n+5}) \\ &&+b^{5}[d_{lb}( \theta_{n+4}, \theta_{n+5})+d_{lb}( \theta_{n+5}, \theta_{n+7})+d_{lb}( \theta_{n+7}, \theta_{n+6})] \\ &\leq &bd_{lb}( \theta_{n}, \theta_{n+1})+b^{2}d_{lb}( \theta_{n+1}, \theta_{n+2})+\left( b+b^{3}\right) d_{lb}( \theta_{n+2}, \theta_{n+3}) \\ &&+\left( b^{2}+b^{4}\right) d_{lb}( \theta_{n+3}, \theta_{n+4})+\left( b^{3}+b^{5}\right) d_{lb}( \theta_{n+4}, \theta_{n+5})+\cdots \\ & \lt &2[bd_{lb}( \theta_{n}, \theta_{n+1})+b^{2}d_{lb}( \theta_{n+1}, \theta_{n+2})+b^{3}d_{lb}( \theta_{n+2}, \theta_{n+3}) \\ &&+b^{4}d_{lb}( \theta_{n+3}, \theta_{n+4})+b^{5}d_{lb}( \theta_{n+4}, \theta_{n+5})+\cdots ] \\ &\leq &2[b\delta _{b}^{n}(d_{lb}( \theta_{0}, \theta_{1}))+b^{2}\delta _{b}^{n+1}(d_{lb}( \theta_{0}, \theta_{1}))+b^{3}\delta _{b}^{n+2}\left( d_{lb}( \theta_{0}, \theta_{1})\right) +\cdots ] \\ & \lt &2[b\delta _{b}(\delta _{b}^{\nu (\varepsilon ^{\prime })-1}(d_{lb}( \theta_{0}, \theta_{1})))+b^{2}\delta _{b}^{2}(\delta _{b}^{\nu (\varepsilon ^{\prime })-1}(d_{lb}( \theta_{0}, \theta_{1})))+\cdots ] \lt 2\varepsilon ^{\prime } = \varepsilon . \end{eqnarray*}$

It follows that

$\begin{equation} \lim\limits_{n,m\rightarrow +\infty }d_{lb}( \theta_{n}, \theta_{m}) = 0. \end{equation}$

(2.4)

Thus $\{ \theta_{n}\}$ is a Cauchy sequence in $(U, d_{lb})$ . As $(U, d_{lb})$ is complete, so there exists $\theta^{\ast }$ in $U$ such that $\{ \theta_{n}\}$ converges to $\theta^{\ast },$ that is,

$\begin{equation} \lim\limits_{n\rightarrow +\infty }d_{lb}( \theta_{n}, \theta^{\ast }) = 0. \end{equation}$

(2.5)

Now, suppose that $d_{lb}(\theta^{\ast }, S \theta^{\ast }) > 0$ . Then

$\begin{eqnarray*} d_{lb}( \theta^{\ast },S \theta^{\ast }) &\leq &b[d_{lb}( \theta^{\ast }, \theta_{n})+d_{lb}( \theta_{n}, \theta_{n+1})+d_{lb}( \theta_{n+1},S \theta^{\ast }) \\ &\leq &b[d_{lb}( \theta^{\ast }, \theta_{n+1})+d_{lb}( \theta_{n}, \theta_{n+1})+d_{lb}(S \theta_{n},S \theta^{\ast }). \end{eqnarray*}$

Since $\alpha (\theta_{n}, \theta^{\ast })\geq 1,$ we obtain

$\begin{eqnarray*} d_{lb}( \theta^{\ast },S \theta^{\ast }) &\leq &bd_{lb}( \theta^{\ast }, \theta_{n+1})+bd_{lb}( \theta_{n}, \theta_{n+1})+b\delta _{b}(\max \{d_{lb}( \theta_{n}, \theta^{\ast }), \\ &&\frac{d_{lb}\left( \theta^{\ast },S \theta^{\ast }\right) .d_{lb}\left( \theta_{n}, \theta_{n+1}\right) }{a+d_{lb}\left( \theta_{n}, \theta^{\ast }\right) },\text{ }d_{lb}( \theta_{n}, \theta_{n+1})\text{ }d_{lb}( \theta^{\ast },S \theta^{\ast })\}). \end{eqnarray*}$

Letting $n\rightarrow +\infty$ , and using the inequalities (2.4) and (2.5), we obtain $d_{lb}(\theta^{\ast }, S \theta^{\ast })\leq b\delta _{b}(d_{lb}(\theta^{\ast }, S \theta^{\ast })).$ This is not true, because $b\delta _{b}(t) < t$ for all $t > 0$ and hence $d_{lb}(\theta^{\ast }, S \theta^{\ast }) = 0$ or $\theta^{\ast } = S \theta^{\ast }$ . Hence $S$ has a fixed point $\theta^{\ast }$ in $U.$

Remark 2.2. By taking fourteen different proper subsets of $D_{lb}(\theta, \nu),$ we can obtainvnew results as corollaries of our result in a complete $r.b.m.$ space with coefficient $b.$

We have the following result without using $\alpha$ -dominated mapping.

Theorem 2.3. Let $(U, d_{lb})$ be a complete $r.b.m.$ space with coefficient $b, \; S:U\rightarrow U$ , $\{ \theta_{n}\}$ be a Picard sequence. Suppose that, for some $\delta _{b}\in \varpi _{b}$ , we have

$\begin{equation} d_{lb}(S \theta,S\nu )\leq \delta _{b}(D_{lb}( \theta,\nu )) \end{equation}$

(2.6)

for all $\theta, \nu \in \{ \theta_{n}\}$ . Then $\{ \theta_{n}\}$ converges to $\theta^{\ast }\in U.$ Also, if (2.6) holds for $\theta^{\ast }$ , then $S$ has a fixed point $\theta^{\ast }$ in $U.$

We have the following result by taking $\delta _{b}\left(t\right) = ct$ , $t\in \mathbb{R} ^{+}$ with $0 < c < \frac{1}{b}$ without using $\alpha$ -dominated mapping.

Theorem 2.4. Let $(U, d_{lb})$ be a complete $r.b.m.$ space with coefficient $b$ , $S:U\rightarrow U$ , $\{ \theta_{n}\}$ be a Picard sequence. Suppose that, for some $0 < c < \frac{1}{b}$ , we have

$\begin{equation} d_{lb}(S \theta,S\nu )\leq c(D_{lb}( \theta,\nu )) \end{equation}$

(2.7)

for all $\theta, \nu \in \{ \theta_{n}\}$ . Then $\{ \theta_{n}\}$ converges to $\theta^{\ast }\in U.$ Also, if (2.7) holds for $\theta^{\ast }$ , then $S$ has a fixed point $\theta^{\ast }$ in $U.$

Ran and Reurings ^[16] gave an extension to the results in fixed point theory and obtained results in partially ordered metric spaces. Arshad et al. ^[3] introduced $\preceq$ -dominated mappings and established some results in an ordered complete dislocated metric space. We apply our result to obtain results in ordered complete $r.b.m.$ space.

Definition 2.5. $(U, \preceq, d_{lb})$ is said to be an ordered complete $r.b.m.$ space with coefficient $b$ if

(ⅰ) $(U, \preceq)$ is a partially ordered set.

(ⅱ) $(U, d_{lb})$ is an $r.b.m.$ space.

Definition 2.6. ^[3] Let $U$ be a nonempty set, $\preceq$ is a partial order on $\theta.$ A mapping $S:U\rightarrow U$ is said to be $\preceq$ -dominated on $A$ if $a\preceq Sa$ for each $a\in A\subseteq \theta.$ If $A = U,$ then $S:U\rightarrow U$ is said to be $\preceq$ -dominated.

We have the following result for $\preceq$ -dominated mappings in an ordered complete $r.b.m.$ space with coefficient $b$ .

Theorem 2.7. Let $(U, \preceq, d_{lb})$ be an ordered complete $r.b.m.$ space with coefficient $b$ , $S:U\rightarrow U, \; \{ \theta_{n}\}$ be a Picard sequence and $S$ be a $\preceq$ -dominated mapping on $\{ \theta _{n}\}.$ Suppose that, for some $\delta _{b}\in \varpi _{b}$ , we have

$\begin{equation} d_{lb}(S \theta,S\nu )\leq \delta _{b}(D_{lb}( \theta,\nu )), \end{equation}$

(2.8)

for all $\theta, \nu \in \{ \theta_{n}\}$ with $\theta \preceq \nu$ . Then $\{ \theta_{n}\}$ converges to $\theta^{\ast }\in U.$ Also, if (2.8) holds for $\theta^{\ast }$ and $\theta_{n}\preceq \theta^{\ast }$ for all $n\in \mathbb{N} \cup \{0\}$ . Then $S$ has a fixed point $\theta^{\ast }$ in $U$ .

Proof. Let $\alpha :U\times U\rightarrow \lbrack 0, +\infty)$ be a mapping defined by $\alpha (\theta, \nu) = 1$ for all $\theta, \nu \in U$ with $\theta \preceq \nu$ and $\alpha (\theta, \nu) = \frac{4}{11}$ for all other elements $\theta, \nu \in U.$ As $S$ is the dominated mappings on $\{ \theta_{n}\},$ so $\theta \preceq S \theta$ for all $\theta \in \{ \theta_{n}\}.$ This implies that $\alpha (\theta, S \theta) = 1$ for all $\theta \in \{ \theta_{n}\}.$ So $S:U\rightarrow U$ is the $\alpha$ -dominated mapping on $\{ \theta_{n}\}.$ Moreover, inequality (2.8) can be written as

$\begin{equation*} d_{lb}(S \theta,S\nu )\leq \delta _{b}(D_{lb}( \theta,\nu )) \end{equation*}$

for all elements $\theta, \nu$ in $\{ \theta_{n}\}$ with $\alpha (\theta, \nu)\geq 1.$ Then, as in Theorem 2.1, $\{ \theta_{n}\}$ converges to $\theta^{\ast }\in U.$ Now, $\theta_{n}\preceq \theta^{\ast }$ implies $\alpha (\theta_{n}, \theta^{\ast })\geq 1.$ So all the conditions of Theorem 2.1 are satisfied. Hence, by Theorem 2.1, $S$ has a fixed point $\theta^{\ast }$ in $U$ .

Now, we present an example of our main result. Note that the results of George et al. ^[11] and all other results in rectangular $b$ -metric space are not applicable to ensure the existence of the fixed point of the mapping given in the following example.

Example 2.8. Let $U = A\cup B,$ where $A = \{\frac{1}{n}:n\in \{2, 3, 4, 5\}\}$ and $B = [1, \infty ].$ Define $d_{l}:U\times U\rightarrow \lbrack 0, \infty)$ such that $d_{l}(\theta, \nu) = d_{l}(\nu, \theta)$ for $\theta, \nu \in U$ and

$\begin{equation*} \left\{ \begin{array}{c} d_{l}(\frac{1}{2},\frac{1}{3}) = d_{l}(\frac{1}{4},\frac{1}{5}) = 0.03 \\ d_{l}(\frac{1}{2},\frac{1}{5}) = d_{l}(\frac{1}{3},\frac{1}{4}) = 0.02 \\ d_{l}(\frac{1}{2},\frac{1}{4}) = d_{l}(\frac{1}{5},\frac{1}{3}) = 0.6 \\ d_{l}( \theta,\nu ) = \left\vert \theta-\nu \right\vert ^{2}\rm{ \ \ \ \ otherwise} \end{array} \right. \end{equation*}$

be a complete $r.b.m.$ space with coefficient $b \; = 4 > 1$ but $\ (U, d_{l})$ is neither a metric space nor a rectangular metric space. Take $\delta _{b}(t) = \frac{t}{10},$ then $b\delta _{b}(t) < t.$ Let $S:U\rightarrow U$ be defined as

$\begin{equation*} S \theta = \left\{ \begin{array}{c} \frac{1}{5}{\rm{ \ \ \ \ \ \ \ \ if }}\; \theta \in A \\ \frac{1}{3}\rm{ \ \ \ \ \ \ \ \ if} \theta = 1 \\ 9 \theta^{100}+85\rm{\ otherwise.} \end{array} \right. \end{equation*}$

Let $\theta_{0} = 1.$ Then the Picard sequence $\{ \theta_{n}\}$ is $\{1, \frac{1}{ 3}, \frac{1}{5}, \frac{1}{5}, \frac{1}{5}, \cdots \}$ . Define

$\begin{equation*} \alpha ( \theta,\nu ) = \left\{ \begin{array}{c} \frac{8}{5}{\rm{ \ \ \ \ \ \ \ \ if}} \theta ,\nu \in \{ \theta_{n}\} \\ \frac{4}{7}\rm{\ \ \ \ \ \ \ \ \ \ \ \ otherwise.} \end{array} \right. \end{equation*}$

Then $S\$ is an $\alpha$ -dominated mapping on $\{ \theta_{n}\}.$ Now, $S$ satisfies all the conditions of Theorem 2.1. Here $\frac{1}{5}$ is the fixed point in $U.$

3. Fixed point results for graphic contractions

Jachymski ^[13] proved the contraction principle for mappings on a metric space with a graph. Let $(U, d)$ be a metric space and $\bigtriangleup$ represents the diagonal of the cartesian product $U\times U.$ Suppose that $G$ be a directed graph having the vertices set $V(G)$ along with $U,$ and the set $E(G)$ denoted the edges of $U_{{}}$ included all loops, i.e., $E(G)\supseteq \bigtriangleup.$ If $G$ has no parallel edges, then we can unify $G$ with pair $(V(G), E(G))$ . If $l$ and $m$ are the vertices in a graph $G,$ then a path in $G$ from $l$ to $m$ of length $N \; (N\in \mathbb{N})$ is a sequence $\{ \theta_{i}\}_{i = o}^{N}$ of $N+1$ vertices such that $l_{o} = l, \; l_{N} = m$ and $(l_{n-1}, l_{n})\in E(G)$ where $i = 1, 2, \cdots N$ (see for detail ^{[7,8,12,14,18,19]}). Recently, Younis et al. ^[20] introduced the notion of graphical rectangular $b$ -metric spaces (see also ^[5,6,21]). Now, we present our result in this direction.

Definition 3.1. Let $\theta$ be a nonempty set and $G = (V(G), E(G))$ be a graph such that $V(G) = U$ and $A\subseteq U$ . A mapping $S:U\rightarrow U$ is said to be graph dominated on $A$ if $(\theta, S \theta)\in E(G)$ for all $\theta \in A$ .

Theorem 3.2. Let $(U, d_{lb})$ be a complete rectangular $b$ -metric space endowed with a graph $G$ , $\{ \theta_{n}\}$ be a Picard sequence and $S:U\rightarrow U$ be a graph dominated mapping on $\{ \theta_{n}\}$ . Suppose that the following hold:

(i) there exists $\delta _{b}\in \varpi _{b}$ such that

$\begin{equation} d_{lb}(S \theta,S\nu )\leq \delta _{b}(D_{lb}( \theta,\nu )), \end{equation}$

(3.1)

for all $\theta, \nu \in \{ \theta_{n}\}$ and $(\theta_n, \nu)\in E(G)$ . Then $(\theta_{n}, \theta_{n+1})\in E(G)$ and $\{ \theta_{n}\}$ converges to $\theta^{\ast }.$ Also, if (3.1) holds for $\theta^{\ast }$ and $(\theta_{n}, \theta^{\ast })\in E(G)$ for all $n\in \mathbb{N} \cup \{0\}$ , then $S$ has a fixed point $\theta^{\ast }$ in $U.$

Proof. Define $\alpha :U\times U\rightarrow \lbrack 0, +\infty)$ by

$\begin{equation*} \alpha ( \theta,\nu ) = \left\{ \begin{array}{c} 1,{\rm{\ if }}\; \theta,\nu \in U,\text{ }( \theta,\nu )\in E(G) \\ \frac{1}{4},\rm{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ otherwise}. \end{array} \right. \end{equation*}$

Since $S$ is a graph dominated on $\{ \theta_{n}\},$ for $\theta \in \{ \theta_{n}\}, \; (\theta, S \theta)\in E(G)$ . This implies that $\alpha (\theta, S \theta) = 1$ for all $\theta \in \{ \theta_{n}\}.$ So $S:U\rightarrow U$ is an $\alpha$ -dominated mapping on $\{ \theta_{n}\}.$ Moreover, inequality (3.1) can be written as

$\begin{equation*} d_{lb}(S \theta,S\nu )\leq \delta _{b}(D_{lb}( \theta,\nu )), \end{equation*}$

for all elements $\theta, \nu$ in $\{ \theta_{n}\}$ with $\alpha (\theta, \nu)\geq 1.$ Then, by Theorem 2.1, $\{ \theta_{n}\}$ converges to $\theta^{\ast }\in U.$ Now, $(\theta_{n}, \theta^{\ast })\in E(G)$ implies that $\alpha (\theta_{n}, \theta^{\ast })\geq 1.$ So all the conditions of Theorem 2.1 are satisfied. Hence, by Theorem 2.1, $S$ has a fixed point $\theta^{\ast }$ in $U$ .

Acknowledgments

The authors would like to thank the Editor, the Associate Editor and the anonymous referees for sparing their valuable time for reviewing this article. The thoughtful comments of reviewers are very useful to improve and modify this article.

Conflict of interest

The authors declare that they have no competing interests.

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